The number of primary alcohols possible with the formula $C_{4} H_{10} O$ is :

2

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3

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4

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5

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Solution

The correct option is D $2$
Primary alcohol means $- {C H}_{2} - O H$
So remaining carbons are $= 3$
Total no. of radical of $3 C$ are $= 2$
$C - C - C -$ and $C - C | C -$
So, total possible primary alc. are $= 2$

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